give some basic information about the field of Civil Engineering, Infrastructural . analysis and design of structural components such as matrix method of. Elements of Civil Engineering and Engineering Mechanics - Free ebook download as PDF File .pdf), Text File .txt) or read book online for free. ELEMENTS OF. F.Y. nbafinals.info [Civil Engineering). CE Elements of Civil Engineering,. Time: 3 Hrs. MAX MARA ati. -. Instruction to candidates: 1. Answer ALL of questions. 2.

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The cement-concrete roads are becoming popular because of the fact that concrete of desired quality can be prepared by modern techniques of cement-concrete construction. The advantages of a cement-concrete road are: The disadvantages are: Classification of Roads by Nagpur Road Plan The classification based on location and function is a more acceptable classification. According to Nagpur plan, five different categories of roads are as follows: National Highways NHs: National highways are the main highways running through the length and breadth of the country, connecting major ports, foreign highways, capitals of large states and large industrial and tourist centres including roads required for strategic movements of troops.

State Highways SHs: State highways are the arterial roads of a state, connecting with the national highways, capitals of adjacent states, district headquarters and important cities within the state. A state highway serves as a connecting link for traffic to and from district roads. Major district roads are the important roads within a district, serving areas of production and market and connecting them with each other or with main highways of a district.

Other district roads are the roads serving the rural areas of production and providing them with an outlet to market centres, Taluka headquarters. Village Roads VRs: Village roads are the roads connecting villages or the groups of villages with each other to the nearest road of higher category.

Components of roads and their functions. Just like other structures, the highway or the road structure is also required to be designed carefully for the traffic load to be carried, physical and geological features, and climatic conditions of the locality and various other factors which would affect the stability and utility of the highway.

For the purpose of design, the road structure may be considered to consist essentially of the following four component parts as shown in Figure 1.

Subsoil 2. Subgrade 3. Base 4. This is the natural or prepared soil on which the road has to be formed. It should be stable and strong to carry safely the traffic load and weight of roadway construction. The subgrade or the formation functions as a support to the road surface and its foundation. The life of the road primarily depends on stable and dry subgrade.

Its level may be same or above or below the natural ground level. The support given to the road structure by the subgrade is an important factor and hence, considerable attention should be paid to the proper preparation of subgrade before the road structure is laid on it.

The base or foundation may consist of two layers, the bottom layer being known as sub-base or soling or bottoming.

The sub-base should be stable and it should be capable of resisting distortion under traffic loads to a great extent. The function of a road base is to transmit the load from the surfacing to the subgrade.

It should possess structural stability and should be of sufficient thickness to develop a good bond with the surfacing.

The topmost layer on which the traffic directly travels is known as road surfacing or wearing layer or wearing course. The main function of road surfacing is to provide a smooth and stable running surfacing which is suitable for the type and intensity of traffic anticipated on the road. The surfacing should be impervious and should protect the base and the subgrade from the.

The desirable qualities of surfacing are durability, stability, non-slipperiness, economical and dustlessness. It should also be able to resist displacement by traffic and should therefore must be well keyed on to the base.

The governing factors which would affect the design of the the above components of road structure are initial cost, availability of local materials, volume and class of traffic, climatic conditions of the locality, etc. Dams A dam is an impervious hydraulic structure constructed across a river to store water on the upstream side.

It acts as a barrier to form a reservoir. Dams may be classified into different categories depending upon the purpose they serve or on the basis of the material used for construction. Classification according to use. Storage dam: A storage dam is constructed to store water on its upstream side during the period of excess supply and use it during the period of scarcity.

Diversion dam: A diversion dam raises the water level slightly in the river, and thus water is diverted into a canal system. A diversion dam is always smaller in height and no reservoir is formed on its upstream to store water. Detention dam: A detention dam is constructed to store water during floods, in order to be able to release water gradually at a safe rate when the flood recedes.

By the provision of a detention dam, damages dowstream due to flood are reduced. Classification according to material. Rigid dam: Rigid dams are those which are constructed of rigid material such as masonry, concrete, steel or timber. Non-rigid dam: Classification according to flow condition. Overflow dam: If water is allowed to flow over the top of the dam, it is known as overflow dam. Such dams are also known as spillways.

Gates are normally provided over spillways for allowing the water to pass downstream during floods and for extra storage of water after the floods. Non-overflow dams: If water is not allowed to flow over the top of the dam, it is known as a non-overflow dam. All the dams are generally a combination of overflow and non-overflow dams. Gravity dams Rigid dams.

A gravity dam is one in which the external forces such as water pressure, silt pressure, wave pressure, etc. Thus the weight of the dam or the gravity forces maintain the stability of the dam. A gravity dam may be constructed either of masonry or concrete. The different terms of importance are as follows: Maximum water level or full reservoir level: The maximum level to which the water rises during the worst flood is known as the maximum water level or full reservoir level.

Minimum pool level: The lowest water surface elevation up to which the water in the reservoir can be used is called the minimum pool level. Normal pool level: It is the maximum elevation to which the reservoir water surface will rise during normal operating conditions. Useful and dead storage: The volume of water stored in the reservoir between the minimum pool level and normal pool level is called useful storage. The volume of water stored in the reservoir below the minimum pool level is known as dead storage.

Free board: The margin between the maximum water level and top of the dam is known as free board. Free board must be provided to avoid the possibility of water spilling over the top of the dam due to wave action.

Drainage gallery: A gallery provided near the foundation to drain off the water which seeps through the foundation and the body of dam is called the drainage gallery. Earth dams and rockfill dams Non-rigid dams. Earth dams are constructed using the locally available soils and gravels; they can be used up to moderate heights only. Earth dams may be classified into: Homogeneous type dams 2. Zoned type dams 3. Diaphragm type dams. Bridges are an integral part of the infrastructure for the economic growth of the country and therefore they cannot remain static in relation to a growing economy.

With the increase in population and growth in industrial and agricultural output, bridges along with railways have to keep pace with the developments through provision of transport facilities in areas where these have been lacking and increased facilities where these are inadequate to meet the demands of the travelling public as well as those of the industrial and agricultural sectors.

The continuous improvement in bridge systems is an important indication of the current scientific and technological revolution which has a direct bearing on the progress of productive forces.

Bridges play a vital role in shaping the history of a country as the existence or otherwise of these bridges very much affect the movement of troops during the hostilities and consequently these bridge points are required to defend ourselves from the onslaught of the invaders.

No road system is considered complete in itself without the provision of such cross drainage works in the form of bridges, so that there is no hindrance to the free flow of traffic, especially during the rains. Many types of bridges have been constructed from the times of olden days ranging from the timber bridges to modern steel and pre-stressed concrete bridges. The modern bridges are usually constructed of steel, being durable and easy to fabricate.

The type of steel bridge to be used depends on the span. Cable stayed bridges are usually constructed over gorges in hilly countries where the water flow in the stream below is very fast. Cable stayed bridges are constructed with boulders where piers cannot be easily constructed. These bridges have a span of metres and are usually meant for pedestrian traffic.

The deck is supported by long cables carried over tall towers provided at their ends. Timber bridges Figure 1. When the materials for constructing permanent bridges are not available or the same are available at long distances entailing huge cost of construction, these wooden bridges are suitable and cheaper. Timber bridges have a life-span of about 15 years and may require replacement thereafter.

They are likely to be destroyed by fire and fungus growth if not properly maintained. They are suitable for small spans of 5 metres or so. The state Jammu and Kashmir has several such wooden bridges constructed many years ago. Bridges constructed purely of timber may have either timber arch construction or be of strutted beam type. The points to be borne in mind while deciding the construction of timber bridges are: Duration of time for which the bridge is required Amount of load required to be carried Quality of timber available Method of erection.

Floating bridges. Floating bridges, also known as pontoon bridges, are temporary structures constructed over rivers in times of emergency during wars where the time is of fundamental importance. The advantage of these floating bridges is that they can be erected very quickly as the component parts of the bridge are of pre-fabricated types as per the dimensions approved by the Indain Road Congress and hence.

The materials used for the parts are strong, durable and non-corrosive. A typical view of a floating bridge is shown in Figure 1. As stated before, bridges are also classified according to the method of giving clearance to navigation. Under that classification, movable bridges are of the following four types, namely: Swing bridges Bascule bridges Lift bridges Transporter bridges Swing bridges: In case of swing bridges there is a disc bearing placed over the central pier on which two pans of continuous trusses are provided Figure 1.

By an arrangement of rotating gear, the whole truss rotates horizontally through 90 when its axis becomes parallel to the direction of flow of water and at that position pedestrians or other vehicles cannot move. Bascule bridges: A rough idea of the working of a bascule bridge can be had from the common toll tax barrier arrangement in India where a full span pole can be lowered and raised with the help of a counterweight placed at one end.

In a bascule bridge, instead of the pole, there is provided a hinged truss which can be raised or lowered in a vertical plane with the help of a counterweight or rack and pinion arrangement or by cables. A general view of single and double bascule bridges is shown in Figure 1.

Lift bridges: For wide channels the bascule bridges are not economical. In such cases the vertical lift bridges are recommended to give clearance to navigation. A vertical lift consists of a roadway truss which is lifted up by means of cables passing over pulleys attached to each of the two tall towers erected at either ends. These cables are connected to counterweights at the other end.

A general view of a vertical bridge is shown in Figure 1. Transporter bridges: In case of transporter bridges, a moving cage is suspended from an overhead truss with the help of a cable or wire ropes.

The overhead truss rests on two towers and it contains rails for the cage to roll. The cage is loaded with persons or goods and it is then allowed. This type of bridge is used within a harbour area to provide an arrangement for shifting of men and materials across a channel. A general view of a transporter bridge is shown in Figure 1. Suspension bridges are long-span bridges involving creative works and great skill on the part of the bridge engineer. For the stability of such long-span bridges of m and above, considerations for the control of aerodynamic movement, the local angle changes in the deck and vertical oscillations caused by serve winds are of vital importance, failing which a failure of the bridge may result.

Cantilever bridges, due to their heavier weights and labour involved in construction compared to the cable stayed bridges for the same spans, are no longer popular.

It is so called as in a single span there are cantilever arms from each pier to the ends where the freely supported suspended span rests. Then there are anchor arms at either end between the abutment and the pier.

Some views of contilever bridges are shown in Figure 1. For rail-road or long railway bridges of spans to meters, truss bridges Figure 1.

There are various types of truss bridges like the Warren truss bridge, Pratt truss and N-truss, but the most common form is the Warren truss type. In case of arch bridges the abutments must be strong enough to carry the thrust exerted by the ribs. In such cases heavy expensive abutments will be required, unless there is a gorge with rocky banks which might be strong enough to bear the load exerted by the ribs; and it is in such situations where arch bridges are most recommended.

Arch bridges may be made of masonry or steel. Masonry arch bridges are not recommended for spans exceeding 6 metres, especially in seismic zones. For long spans, steel arch bridges are generally used. The arch bridges may be rigid, of two hinge or three hinge type. See Figure 1. Briefly explain the different fields of civil engineering or scope or civil engineering. What is infrastructure? Explain the impact of infrastructural development on a country. Briefly explain the role of civil engineers in the infrastructural development.

Explain the classification of roads. Write a brief note on classification of bridges. What is dam? Briefly describe the different types of dams. Engineering mechanics is the application of mechanics to the solution of engineering problems. It is broadly classified into three types: Mechanics of Rigid Bodies It is the branch of science which deals with the study of bodies that do not undergo any deformation under the application of forces.

It can further be classified into Statics and Dynamics. It is the branch of mechanics which deals with the study of the behaviour of bodies or particles in the state of rest. It is the branch of mechanics which deals with the study of the behaviour of bodies or particles in the state of motion. Dynamics is further divided into two types: The forces causing the motion are not considered. The forces causing the motion are mainly considered. Mechanics of Deformable Bodies It is the branch of science which deals with the study of bodies that undergo deformation under the application of forces.

It is classified into: Mechanics of Fluids It is the branch of science which deals with the study of fluids. Fluids can be classified into: The assemblage of a number of particles is known as a body. Rigid body: A rigid body is one in which the positions of the constituent particles do not change under the application of external forces, such as the position of particles 1 and 2 in Figure 2.

Deformable body: A deformable body is one in which the positions of constituent particles change under the application of external forces, such as the positions of particles 1 and 2 in Figure 2. Mass m: The total amount of matter present in a body is known as its mass. The unit of mass is the kilogram, abbreviated kg. A body is attracted towards the earth due to gravitation.

This causes an acceleration directed towards the centre of earth. It is called acceleration due to gravity and is denoted by g. The resulting force is equal to the weight of body. A physical quantity which has only magnitude, is called scalar quantity.

For example, time, mass, density, volume, distance, and so forth. Vector quantity: A physical quantity which has a direction in addition to magnitude, is known as vector quantity. For example, force, displacement, velocity, acceleration, and so forth. A continuous distribution of molecules in a body without intermolecular space is called the continuum. Newtons Laws of Motion Newtons First law: This law states that every body continues in its state of rest or of uniform motion along a straight line, so long as it is under the influence of a balanced force system.

Newtons Second law: This law states that the rate of change momentum of a body is directly proportional to the impressed force and it takes place in the direction of the force acting on it. Newtons Third law: It is the external agency which tends to change the state of a body or a particle.

When a force is applied to a body which is at rest, the body may remain in the state of rest or it may move with some velocity. The SI unit of force is newton. Elements of a Force or Characteristics of a Force A force can be identified by its four characteristics: Figure 2.

The direction of a force can be represented by an arrowhead. It is the line along which the force acts. It is the point at which the force acts. Point force. A force which is acting at a fixed point is known as the point force. Let us consider a man climbing a ladder. The weight of the man is not actually concentrated at a fixed point but for the purpose of analysis it is assumed to be concentrated at a particular point. If two or more forces are acting on a body or a particle, then it is said to be a force system, such as that shown in Figure 2.

Types of Force System The types of force system are: Coplanar force system 2. Non-coplanar force system 3. Collinear force system. Coplanar force system. If two or more forces are acting in a single plane, then it is said to be a coplanar force system.

The types of coplanar force system are: If two or more forces are acting in a single plane and their lines of action pass through a single point, then it is said to be a coplanar concurrent force system. See Figure 2. If two or more forces are acting in a single plane and their lines of action do not meet at a common point, then the forces constitute a coplanar non-concurrent force system.

If two or more forces are acting in a single plane with their lines of action parallel to one another, then it is said to be a coplanar parallel force system. The coplanar parallel force system is of two types: All the forces act parallel to one another and are in the same direction, as shown in Figure 2. The forces act parallel to another, but some of the forces have their line of action in opposite directions, as shown in Figure 2.

If two or more forces are acting in different planes, the forces constitute a non-coplanar force system. Such a system of forces can be i Non-coplanar concurrent force system ii Non-coplanar non-concurrent force system iii Non-coplanar parallel force system. If a system has two or more forces acting on different planes but pass through the same point, then it is said to be a non-coplanar concurrent force system.

If two or more forces are acting on different planes but do not pass through the same point, they constitute a non-coplanar non-concurrent force system. If two or more forces are acting in different planes and are parallel to one another, the system is said to be a non-coplanar parallel force system. If the lines of action of two or more forces coincide with one another, it is called a collinear force system as shown in Figure 2. If the lines of action of the forces do not coincide with one another, it is called a non-collinear force system as shown in Figure 2.

Principle of Transmissibility of Forces This principle states that a force can be transmitted from one point to another point along the same line of action such that the effect produced by the force on a body remains unchanged.

Let us consider a rigid body subjected to a force of F at point O as shown in Figure 2. According to the principle of transmissibility, the force F can be transmitted to a new point O along the same line of action such that the net effect remains unchanged.

Principle of Superposition of Forces This principle states that the net effect of a system of forces on a body is same as that of the combined effect of individual forces on the body. Principle of Physical Independence of Forces This principle states that the action of a force on a body is not affected by the action of any other force on the body.

Resolution of a Force The process of splitting of a force into its two rectangular components horizontal and vertical is known as resolution of the force, as shown in Figure 2. In this figure, F is the force which makes an angle q with the horizontal axis, and has been resolved into two components, namely Fx and Fy, along the x-axis and y-axis respectively. If the force F makes an angle of q with the horizontal, the horizontal component of the force is F cos q.

Composition of Forces It is the process of combining a number of forces into a single force such that the net effect produced by the single force is equal to the algebraic sum of the effects produced by the individual forces. The single force in this case is called the resultant force which produces the same effect on the body as that produced by the individual forces acting together.

For example, in Figure 2. The positive and negative convention of forces used in the resolution of forces in Figure 2. The resultant of a system of coplanar concurrent forces can be determined by the following methods. Parallelogram law: If two forces are acting simultaneously on a particle and away from the particle, with the two adjacent sides of the parallelogram representing both the magnitude and direction of forces, the magnitude and direction of the resultant can be represented by the diagonal of the parallelogram starting from the common point of the two forces.

Triangle law: If two forces acting simultaneously on a particle can be represented both in magnitude and direction by the two sides of a triangle taken in order, then the magnitude and direction of the resultant can be represented by the third side of a triangle, taken in opposite order. This is illustrated in Figure 2. Polygon law: If a number of forces acting on a particle can be represented in both magnitude and direction by the sides of the polygon taken in order, then the resultant can be represented in magnitude and direction by the closing side of the polygon taken in the opposite order.

Moment of a Force The turning effect produced by a force on a body is known as the moment of the force. The magnitude of the moment is given by the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the point or axis of rotation.

This is shown in Figure 2. Types of moments. This is also known as the principle of moments. The theorem states that the algebraic sum of the moments of individual forces of a force system about a point is equal to the moment of their resultant about the same point.

Let R be the resultant of forces P1and P2 and B be the moment centre. Let d, d1 and d2 be the moment arms of forces R, P1 and P2, respectively, from the moment centre B Figure 2. Join AB and consider it as the y-axis and draw the x-axis at right angles to it at A. Let q be the angle made by R with the x-axis and note that the same angle is formed with the y-axis by the perpendicular to R from B and note this point as B1. Types of couple.

Define the following terms: What are the two divisions of dynamics? What are the four characteristics of a force? What are the three types of force systems? Define the principle of superposition of forces. What are the various methods of finding the resultant of a system of coplanar concurrent forces?

Define the moment of a force. Define Varignons theorem of moments. What is a couple? What are two types of couple? In a coplanar concurrent force system, we can calculate the magnitude and direction of the resultant.

The position, however, cannot be determined because all forces are meeting at a common point. Thus, The magnitude of resultant, R. The steps to solve problems in the coplanar concurrent force system are, therefore, as follows: Calculate the algebraic sum of all the forces acting in the x-direction i.

SFx and also in the y-direction i. Determine the magnitude of the resultant using the formula, R. Determine the components of this force along the x and y directions. VTU Jan Solution. Example 3. Determine the resultant in magnitude and direction. Solution Here: VTU March One of the forces is unknown and its magnitude is shown by P.

The resultant has a magnitude of N and is acting along the x-axis. Determine the unknown force P and its inclination with the x-axis. Determine the other force. VTU August Solution Let P be the unknown force, which makes an angle q with the horizontal. If the angle 30 between the two 4 kN forces is fixed; find the angle q.

Also determine the magnitude of the resultant. Determine the direction of the force N such that the hook is pulled in the x-direction. Determine the resultant force in the x-direction. Determine the magnitude and direction of the resultant force. The resultant of the two forces is N. Determine q between the forces and the direction of the resultant. Take the co-ordinate directions as shown in the figure.

Find the components of the force along the horizontal and vertical axes. VTU January Solution. VTU January Determine the effect of N force at O. Solution Apply equal and opposite forces of N at O as shown in Figure 3. The effect produced as shown in Figure 3. Solution Apply equal and opposite forces of 40 kN at O parallel to the given force. See Figure 3.

Determine the resultant of the four forces acting on a particle as shown in Figure 3. Determine the resultant of the coplanar concurrent force system shown in Figure 3. Determine the resultant of the force system shown in Figure 3. The force system as shown in Figure 3. Compute the value of force P and its inclination with x-axis. The following forces, as shown in Figure 3. Determine the resultant. The two forces P and Q are acting on a bolt at A, as shown in Figure 3.

Determine the magnitude and direction of the resultant. Figure 3. The car is moving along AC. A collar which may slide on a vertical rod is subjected to three forces as shown in Figure 3. If two or more forces are acting in a single plane, but not passing through the single point, such a force system is known as coplanar non-concurrent force system. In a coplanar non-concurrent force system, we can calculate the magnitude, direction and position of the resultant force as follows: Magnitude of resultant,.

The position of the resultant means the calculation of d, or x and y intercepts as shown in Figure 4. In a coplanar non-concurrent force system, the magnitude, direction and position of resultant can be determined. The position of resultant can be determined by using the Varignons theorem or using. Example 4. Compute the magnitude, direction and line of action of the resultant of the system with reference to the point A. The line of action of kN is directly passing through the point A, therefore, the moment produced by the kN force about A is zero.

Solution Convert the uniformly distributed load UDL into point load, i. Determine the magnitude, direction and the line of action of the resultant. Here we can take moment about any point. Determine the magnitude and line of action of single resultant of the system. If the resultant is to pass through B, what should be the magnitude and direction of a couple?

A couple N-m is acting on a bracket when the resultant is at a distance of 1. Each grid is of size 10 mm 10 mm. Determine the x and y intercepts also. VTU February Find the resultant of these forces, and b locate the point where the line of action of the resultant intersects the edge AB of the plate. Locate the resultant force with respect to point D. Magnitude of resultant R: VTU July Out of them, three are shown in the figure. Find the missing force and its position.

Find the resultant of the force shown in Figure 4. Find the magnitude, direction and position of the resultant of the force system, shown in Figure 4.

Four forces are acting on 28 cm 15 cm plate as shown in Figure 4. Find the resultant of these forces. Locate the two points where the line of action of the resultant intersects the edge of the plate. Find the resultant of the system. Find the distance of D from C. Find the resultant of the force system shown in Figure 4.

Also, find the magnitude and sense of a single vertical force to be applied at A, so that the resultant of entire system passes through B and find the resultant.

First find the resultant and then apply a vertical force of P at A assume down or up , such that the x-intercept is cm. Three forces act on a vertical pole as shown in Figure 4. Find the resultant of the three forces. Locate the point where the resultant cuts the pole.

Various forces to be considered for the stability analysis of a dam are shown in Figure 4. The dam is safe if the resultant force passes through middle-third of base.

Forces acting on 1m length of a dam are shown in Figure 4. Neglecting the weight of the dam, determine the resultant force acting on the dam. Find the point of intersection of the resultant with the base of the dam. Principle of equilibrium. According to this principle, A body is said to be in equilibrium if the algebraic sum of all forces acting on the body is zero, and also if the algebraic sum of moments of forces about any fixed point is zero, i.

Conditions of equilibrium for different force systems. See Figure 5. Sometimes the resultant of the force system is not equal to zero.

That means the body is not in equilibrium. The force which is required to keep the body in equilibrium, is known as equilibrant. Lamis theorem. This theorem states that if three forces acting at a point are in equilibrium, then each force is directly proportional to the sine of angle between the other two forces.

See figure 5. Let us consider a spherical ball of mass m, placed on a horizontal plane and tied to the plane by a string as shown in Figure 5. Figure 5. In Figure 5.

Analyse the given problem by applying the above conditions of equilibrium or by applying the Lamis theorem. Lamis theorem can be applied if only three forces are acting at a point.

Solution Apply two conditions of equilibrium to calculate the magnitude and direction of the unknown force. Example 5. Determine the minimum required magnitude of force F. Also, find the direction of force F. Applying Lamis theorem Figure 5. Determine the magnitude and the direction of the minimum force F that should be exerted at the free end of the rope. F 3 sin Its free-body diagram is shown in Figure 5.

Using Lamis theorem Figure 5. A force of 30 kN is applied at C. Determine the forces in the cables CA and CB. TAC sin Also, find the tensile forces developed in the different segments of the rope. The weights rest with portions AB and CD inclined at angles of 30 and 60 respectively with the vertical. The inclination of BC with vertical is Determine the tensions in the various segments of the cable.

At D Figure 5. Using Lamis theorem, we have TCD sin Also, find the value of W1 and W2, if the portion BC is horizontal.

If all contact surfaces are smooth, determine the reaction at contact surfaces:. Solution At contact points, reaction is developed which is perpendicular to each plane. Suppose the two planes are perpendicular to each other, if one plane makes an angle of with the horizontal, the perpendicular plane makes on angle q with the vertical. Applying Lamis theorem, we have RB sin The weights are N and N respectively.

Find the reactions at all the points of contact.

Therefore it is necessary to consider the FBD of sphere 2 first. Consider the FBD of sphere 2. Assuming that all contact surfaces are smooth, determine the reactions at contact points A, B, C, and D. Solution Consider the FBD of sphere 2: The radii of spheres 1 and 2 are, respectively, 20 mm and 30 mm.

RC RD 30 9. Calculate the reactions at all the points of contact. It is given that: RN 4 RL sin Solution Two equal and opposite reactions will be developed between the contact surfaces of two bodies, as shown in Figure 5. Neglecting the weight of the connecting bar, find the force P applied such that the system is in equilibrium. A ball of weight N rests in a right-angled groove as shown in Figure 5.

The sides of the groove are inclined at an angle of 30 and 60 to the horizontal. If the entire surface is smooth, determine the reactions RA and RC at the points of contact. A circular roller of radius 5 cm and weight N rests on a smooth horizontal surface and is held in position by an inclined bar AB of length 10 cm as shown in Figure 5.

A horizontal force of N is acting at B. Find the tension force in the bar AB and reaction at C. A pendulum ball weighs 50 N. It is pulled sideways by a horizontal force of 20 N.

Calculate the angle of the rope of the pendulum to the vertical and tension in the string as shown in Figure 5. Two spheres each of weight N and of radius 25 cm rest in a horizontal channel of width 90 cm as shown in Figure 5. Find the reactions on the points of contact A, B, C and D. Two identical rollers each of weight N are supported by an inclined plane and vertical wall as shown in Figure 5.

Find the reaction exerted by the wall and the inclined plane at C, D, and F. Two spheres are resting in a trench as shown in Figure 5. Determine the reactions acting on the spheres at A, B, C, and D, assuming the surface to be smooth. Two smooth spheres rest between two vertical columns as shown in Figure 5. The larger and smaller spheres weigh respectively N and N. The diameters of the larger and smaller spheres are 36 cm and 16 cm respectively.

Find the reactions at 1, 2, 3 and 4. Two cylinders are placed as shown in Figure 5. Neglecting friction, find the reactions at all contact surfaces. Find the force S in the string AB and the pressure produced on the floor at the point of contacts D and E. Determine the magnitude and direction of force P such that the force system is in equilibrium.

A beam 20 m long supports a load of 12 kN, the cable AC is horizontal and 10 m long. Find the stress in the cable for the beam shown in Figure 5. What axial forces does the vertical load 16 kN induce in the tie rod and in the jib of the crane shown in the Figure 5.

Neglect the weight of the member. The tie rod will be under tension and the jib under compression. A body weighing 16 kN is supported from a fixed point by a string 10 cm and is kept at rest by a horizontal force P at a distance of 6 cm from the vertical line drawn through the point of suspension.

Find the tension in the string and the value of P Figure 5. A body of weight 70 kN is supported by two strings whose lengths are 6 cm and 8 cm from two points in the same horizontal level. The horizontal distance between the two points is 10 cm. Determine the tensions in the string Figure 5.

At C, weight of N is a suspended. The pulley at E is frictionless. The structural members exert forces on supports known as action.

Similarly, the supports exert forces on structural members known as reaction. A beam is a horizontal member, which is generally placed on supports. The beam is subjected to vertical forces known as action. Supports exert forces, known as reaction, on the beam. Types of Supports The following types of supports are found in practice: Simple supports 2. Roller supports 3. Hinged or pinned supports 4.

Fixed supports Simple supports. Simple supports Figure 6. They restrict translation of the body in one direction only, but do not restrict rotation. Roller supports Figure 6. They restrict translation of the body along one direction only, and rotation is allowed.

Hinged supports Figure 6. Therefore, hinged supports restrict translation in both directions. But rotation is possible. Fixed supports Figure 6. Fixed supports develop an internal moment known as restraint moment to prevent the rotation of the body. It is a beam which consists of simple supports Figure 6. Such a beam can resist forces normal to the axis of the beam.

It is a beam which extends beyond support s. In Figure 6. The overhang portion is BC. A load which is distributed uniformly along the entire length of the beam is known as uniformly distributed load such as the load 20 kN per metre UDL , length shown in Figure 6. A load which varies with the length of the beam is known as uniformly varying load Figure 6. The magnitude of the point load corresponding to a uniformly varying load such as that shown in Figure 6.

In a coplanar non-concurrent force system, three conditions of equilibrium can be applied, namely as follows: Draw the free body diagram of the given beam by showing all the forces and reactions acting on the beam. Apply the three conditions of equilibrium to calculate the unknown reactions at the supports. Example 6. Find the reactions at A and B. Solution It is a coplanar non-concurrent force system, therefore, it is possible to apply three conditions of equilibrium to calculate the reactions at the supports A and B.

Determine the reactions at A and B. The moment at A is anticlockwise in nature. Load is acting at the CG of the triangle as shown in Figure 6. Using the conditions of equilibrium shown in Figure 6. First, we convert the triangular load into a point load as follows: VTU January Solution or. Apply the conditions of equilibrium to calculate the reactions Figure 6. Solution The radius of the roller is negligible. Consider the beam BE Figure 6.

Consider the beam AC Figure 6. Applying the conditions of equilibrium to Figure 6. Calculate the support reactions for the beam loaded and supported as shown in Figure 6. Determine the support reactions for the beam supported and loaded as shown in Figure 6. Determine the support reactions for a beam loaded as shown in Figure 6.

A horizontal beam 6 m long is supported on a knife edge at its end B and the end A, rests on a roller support placed on an inclined plane, having an inclination of as shown in Figure 6.

Find the reactions at the supports A and B. Find the magnitude of anticlockwise couple M to be applied at D so that reaction at F will be 35 kN upwards. Also find the reaction at support B of beam as shown in Figure 6.

Find the support reactions for a beam loaded and supported as shown in Figure 6. Compute the reactions at the supports of beam ABCD which is loaded and supported as shown in Figure 6. A beam ABF is loaded and supported as shown in Figure 6. Find the direction and magnitude of the couple to be applied at D so that the reaction at F will be 50 kN upwards. Also compute the reaction at the F support. They are: Plane trusses are those structures in which all members are lying in a single plane.

Plane trusses are made of several bars or members connected together at the joints by riveting or welding to form a rigid formwork, and also support stationary loads or moving loads. Fluid Mechanics and Hydraulics It is a branch of science in which the study of fluids, i. Usually, the liquid is water and the subject is titled hydraulics. When the water is as rest, the forces exerted by the water on immersed areas are found out by the laws of mechanics.

Thus, the knowledge of these forces is useful in the design of the gates used to control the flood water in case of dams. When water is drawn off from a reservoir and conveyed through closed conduits or open channels, the knowledge of the behaviour of liquids in motion is useful here. Thus in the design of water supply distribution systems, the study of fluid mechanics helps to solve the problems encountered in the design.

Certain machines which work on the water are used for the generation of electricity and are termed hydraulic machines. Knowledge of fluids mechanics is useful in designing these machines so that they give the best possible output. Transportation Engineering This subject deals with the transport of men and materials through different communication routes such as land, water and air.

The railways and roads are the important modes of communication by land. The water transport is feasible only where the rivers, canals are navigable or where the sea coast is available. Transportation by air routes is also increasing day by day. The knowledge of surveying and levelling is very useful before deciding the alignments of roads or railway.

The preparation of contour plan of a hilly region obtained from a levelling operation is useful in deciding the alignment of hill roads. The knowledge of surveying and levelling enables the engineer to decide the alignment of tunnels, which become necessary when the road or railway transportation is to be done through the hilly regions.

Water Resources Engineering Water is such a commodity that it is vital for the existence of mankind. Human beings, animals and plants require water for their survival. Surface water is easy and economical to harness, however, its availability cannot be relied upon continuously since it varies with the season. Water Resources Engineering can be defined as the science which deals with the subject of tapping water either from the surface or subsurface sources.

A water resource is such a vast subject that it includes in itself hydrology, irrigation, hydraulics and water supply. Tremendous volume of water is stored in the earths crust. According to one estimate the total volume stored under the surface of earth may be about 80 million km cube, half of which may be at depths less than m. The use of surface water for irrigation is likely to create problems like waterlogging in certain areas.

Groundwater is obtainable all the year round and its use along with the surface water keeps the subsoil water level within reasonable limits. The judicious use of water for purposes of irrigation has gained such importance in the recent times that water management has become a science in itself.

Engineers have learnt to tame the water resources by construction of dams, construction of bore wells and construction of hydroelectric plants for the benefit of mankind.